A346924 Expansion of e.g.f. 1 / (1 + log(1 - x)^5 / 5!).

1, 0, 0, 0, 0, 1, 15, 175, 1960, 22449, 269577, 3430790, 46480830, 671260876, 10329270952, 169125055736, 2940784282800, 54182845939104, 1055291277366108, 21674715826211532, 468366193441002564, 10624074081842024496, 252432685158931968768, 6270222495850552958004

Offset: 0

Author

Ilya Gutkovskiy, Aug 07 2021

Keywords

• nonn

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..444

Crossrefs

Cf. A007840, A346921, A346922, A346923.

Cf. A000482, A347004, A353200.

Programs

Mathematica

nmax = 23; CoefficientList[Series[1/(1 + Log[1 - x]^5/5!), {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] Abs[StirlingS1[k, 5]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 23}]

PARI

my(x='x+O('x^25)); Vec(serlaplace(1/(1+log(1-x)^5/5!))) \\ Michel Marcus, Aug 07 2021

PARI

a(n) = sum(k=0, n\5, (5*k)!*abs(stirling(n, 5*k, 1))/120^k); \\ Seiichi Manyama, May 06 2022

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * |Stirling1(k,5)| * a(n-k).

a(n) ~ n! * 2^(3/5) * 3^(1/5) * exp(2^(3/5)*15^(1/5)*n) / (5^(4/5) * (exp(2^(3/5)*15^(1/5)) - 1)^(n+1)). - Vaclav Kotesovec, Aug 08 2021

a(n) = Sum_{k=0..floor(n/5)} (5*k)! * |Stirling1(n,5*k)|/120^k. - Seiichi Manyama, May 06 2022